text:
Do all 737NG have inerting systems?
My understanding is that Boeing 737-600/-700/-800/-900 are all 737NG.
I've read before that the previous generation (737 Classic) can't be used in the USA as they don't have fuel inerting systems.
But do all (or nearly all) 737NG have fuel inerting systems? I'm thinking here about 737NG worldwide, and not necessarily those in the USA.

A few classic 737's do still fly in the US. The statement about a lack of fuel inerting being a reason to ground them is incorrect. There are other things that make them expensive to keep flying, but inerting isn't one of them.

The Boeing 737 NG series first flew in 1997, but the fuel inerting system was only certified in 2006 according to b737.org.uk (emphasis mine):

Centre Fuel Tank Inerting
To date, two 737's, 737-400 HS-TDC of Thai Airways on 3 Mar 2001 and 737-300 EI-BZG operated by Philippine Airlines on 5 Nov 1990 have been destroyed on the ground due to explosions in the empty centre fuel tank. The common factor in both accidents was that the centre tank fuel pumps were running in high ambient temperatures with empty or almost empty centre fuel tanks.
Even an empty tank has some unusable fuel which in hot conditions will evaporate and create an explosive mixture with the oxygen in the air. These incidents, and 15 more on other types since 1959, caused the FAA to issue SFAR88 in June 2001 which mandates improvements to the design and maintenance of fuel tanks to reduce the chances of such explosions in the future. These improvements include the redesign of fuel pumps, FQIS, any wiring in tanks, proximity to hot air-conditioning or pneumatic systems, etc.
737s delivered since May 2004 have had centre tank fuel pumps which automatically shut off when they detect a low output pressure and there have been many other improvements to wiring and FQIS. But the biggest improvement will be centre fuel tank inerting. This is universally considered to be the safest way forward, but is very expensive and possibly impractical. The NTSB recommended many years ago to the FAA that a fuel tank inerting system be made mandatory, but the FAA have repeatedly rejected it on cost grounds.
Boeing has developed a Nitrogen Generating System (NGS) which decreases the flammability exposure of the center wing tank to a level equivalent to or less than the main wing tanks. The NGS is an onboard inert gas system that uses an air separation module (ASM) to separate oxygen and nitrogen from the air. After the two components of the air are separated, the nitrogenenriched air (NEA) is supplied to the center wing tank and the oxygenenriched air (OEA) is vented overboard. NEA is produced in sufficient quantities, during most conditions, to decrease the oxygen content to a level where the air volume (ullage) will not support combustion. The FAA Technical Center has determined that an oxygen level of 12% is sufficient to prevent ignition, this is achievable with one module on the 737 but will require up to six on the 747.
The Honeywell NGS was certified by the FAA on 21 Feb 2006 after over 1000hrs flight testing on two 737-NGs. Aircraft have been NGS provisioned since l/n 1935 and it has been installed since l/n 2620 onwards. The NGS requires no flight or ground crew action for normal system operation and is not dispatch critical.

So all older NGs were built without an inerting system. According to this article, the FAA required new models since 2008 to have an inerting system and retrofitting of all 737s until 2018 (emphasis mine):

The headaches are hitting Boeing 737 operators, which make up the bulk of the early adopters among those required to comply with the FAA’s Fuel Tank Flammability Reduction (FTFR) rule. The rule, passed in 2008, called for new-production versions of the affected models—737s, 747s, 757s, 767s, 777s, Airbus A300s/A310s, A320-family models, A330s and A340s—to have fuel tank inerting or flammability reduction systems installed starting in late December 2010. It also required airlines to retrofit half of their affected fleets with an approved system by 2015 and complete the rest by 2018. The phased compliance led operators to prioritize their largest subfleets to ensure they met the first deadline, putting 737s at the top of the list.

It is quite possible that not every airline worldwide has completed this retrofitting.

To follow on from your statement that it's quite possible that not every airline worldwide has completed this retrofitting, I haven't been able to find any indication that Australia (just to name one country) ever mandated any retrofitting.

@user1205197 Yes, me neither. And since most 737s are not crossing continents, I guess the ones not flying to the US are not retrofitted. I also could not find anything definitive for EASA, although they did evaluate it after the FAA changed regulations.

authors: user1205901 - Слава Україні


text:
Is there a KPI to measure average hours flown?
I am working on a report for a small charter airline and I want to set a key performance indicator (KPI) to measure average hours flown.
Is there a practice in aviation field to consider optimum flight hours for an airplane? An arbitrary example: Let's say I have a Boeing 767 and I want to consider if that particular model is profitable (I know it depends on payload, routes, and many other things), does it have to make an average 300 hours per month?
Is there such thing in aviation?

Airplanes don't make money sitting on the ground, if you are asking for a "break even" point considering all costs for that aircraft it may be a very complicated answer because differing labor costs, maintenance, and even the age of the aircraft.

Like you say, it depends on quite a few parameters. You would set up a cash flow Excel sheet with cash in/cash out per month:
Costs

Lease
Salaries
Operational costs including certification, licensing
Office/overhead
Fuel
Airport fees
Maintenance/consumables not covered by lease
Marketing.advertising

Revenue from ticket sales

Number of flights per day
Passenger load factor
Seasonal variations
Market price per ticket
Predicted market share

The companies that lease out aircraft would have filled out scenarios for you. Boeing and Airbus have departments that advise prospective customers on setting up airlines.

i like your answer as well, but i can only chose one accepted answer so i chose  @ymb1's answer. That being said, thank you for pointing me costs and revenue values

No worries, glad to be of help.

Is there a practice in aviation field to consider optimum flight
  hours for an airplane?

Im not really sure what you mean by optimum hours for an air frame but you could likely figure out a predicted up-time (usable and subsequently profitable) for an aircraft. You can start by assuming the aircraft is in the air all the time then subtract for the following,

Time to and from the gate
Time at the gate
If you are in a cold environment - de-ice and other similar pre-flight things
Scheduled maintenance
Some form multiplier for unscheduled maintenance issues (this number is easier to compute after some time of operation.
Time sitting un-used on the ground as there is no route or booking for the plane
Paying everyone involved (maintenance crew, pilots, flight crew etc)

small charter airline

For this you need to look at things like smaller turbo props as they have very different cost breakdowns than a 767.
If you are here in the USA you also need to identify if you are planning to operate a 135 or 121 airline as the governing and rules are somewhat different which will impact cost.

Did you mean cabin crew not flight crew in your last bullet point? Flight crew typically includes the pilots, plus e.g. flight engineer, but cabin crew is commonly discussed separately...

"optimum flight hours" is keeping the aircraft in the air except for mandatory maintenance.
In the 1940-1950 era a DC3 was considered to have a "good" route if it was in the air 2-3 out of 24hrs.
Today, airlines keep aircraft out of the gate closer to 16 out of 24hrs.
Total Airline Operating Cost Breakdown
======================================
44% is operating, includes fuel, direct maintenance, depreciation, and crew
29% is servicing expense, aircraft, traffic, passenger
14% is reservations and sales expense
13% is overhead expense, adds and admin

Cost per block-hour of operations
==================================
B757-200 (avg. 186 seats)
Based on 1252 mile average stage length and 11.3 block-hr daily utilization
CREW $  489
FUEL $  548
MAINTENANCE $  590
OWNERSHIP $  923
--------------------------
TOTAL $ 2,550 per block-hr

Typical costs per block hr
A/C        Seats   blk/Hr   seat/hr
===================================
DC9-30     100     $1,973    $19.73
A320       148     $2,270    $15.33
B727-200   150     $2,555    $17.03
B757-200   186     $2,550    $13.71
B747-400   375     $6,455    $17.21

Source

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text:
In a Practice Forced Landing (PFL), what are Low Key and High Key and how are their locations determined?
Please explain what are "low key" and "high key" in practice forced landings? Are they any particular locations with respect to the airfield? Are they the same for every aircraft? Will they change based on runway elevation?

I've never encountered this term before, where did you see it?

I understand these are used in the PFL techniques practiced in military training, involving a smooth arc from downwind to final rather than the two right angles practiced in PPL training.

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text:
how does recovering a plane from stall lead to exceeding the critical mach speed?
Recovering from stall itself states that the aircraft is below stall speed. So how does recovering from that stall lead to exceeding critical mach.

Providing context will make the question better able to be answered. Recovering a J-3 from a stall has zero chance of exceeding any critical Mach number.  Recovering a T-38, probably likewise, though for entirely different reasons. What "states" what you mention in the question?

just to be pedantic, but stalling does not imply you are below stall speed (S1g). It's linked to the AoA more than anything..

There is no such thing as stall speed. You can stall at mach 1 if you want, all depending on the angle of attack.

There are two types of stall: One is when speed drops below minimum speed, the lift curve slope flattens or inverses, flow separates and lift drops below weight. This is explained here.
The second type is a high speed stall and is caused by shock-induced flow separation when the aircraft approaches Mach 1. In this case, the aircraft will accelerate to a higher flight Mach number where its maximum lift coefficient is smaller and causes lift to drop below weight. Here stall is caused by a speed increase, hence the name.

Lift coefficient variation over Mach number (picture source)
In the coffin corner, the highest point in the envelope of subsonic high performance aircraft, both speeds coincide. Now the aircraft can fly only at one speed, both a speed decrease and a speed increase will cause stall. When the plane recovers from a low speed stall while flying near the coffin corner, it needs to speed up and consequently runs the risk of accelerating beyond the critical Mach speed at which the lift coefficient drops and causes a high speed stall. Recovery is only possible by sinking into higher density and warmer air where the difference between both stall speeds widens and more lift can be created.
Generally, this happens to jets with straight or moderately swept wings which are powerful enough to fly high and fast but are not designed to exceed Mach 1. The Lockheed U-2 would be a prime example.

"It needs to speed up and consequently runs the risk of accelerating beyond the critical Mach speed at which the lift coefficient drops and causes a high speed stall." can you please explain me these lines. How come it would touch the critical mach speed soon while increasing thrust(speed)

@Meghashyam: Did the addition of the plot help in any way?

@Meghashyam I think it's because recovery from a low speed stall includes a pitch forward to regain airspeed quickly. So if you overshoot the speed it can then stall again from the excessive mach. Seems like mach tuck could then create a positive feedback loop. Am I reading that correctly Peter?

How would warmer air aid in recovery? The air in the troposphere will typically increase in temperature as altitude decreases and density increases. Is the increased energy of warmer air that beneficial in exciting the boundary layer?

@TomMcW: Yes, correctly. The lower maximum lift at higher Mach will lock the aircraft into a shallow dive in which it needs less lift (we had that discussion before, remember?) but is prevented from deceleration by the forward component of gravity. It settles at a speed in which drag and that forward acceleration (plus thrust) balance. When it sinks into higher density air, lift picks up and allows it to decelerate again.

@JonathanWalters: No, much simpler. Warmer air has a higher speed of sound, so the Mach number is lower for the same air speed.

@TomMcW How does pitch forward help to  regain airspeed quickly during a stall. Pitch forward is the nose up right it reduces the airspeed know. Correct me if i am wrong.

@Meghashyam By pitch forward I mean nose down, i.e. push the yoke forward

Stall can happen in cruise, at the absolute ceiling. The aircraft stalls at 40,000 ft or higher and at a speed close to cruise, like what happened with AF 447. In that case the aircraft was in a fully developed stall and falling to the ground in a more or less level attitude, at terminal velocity.
Recovery would have been initiated with nose down command, and then the aircraft starts to dive nose down, picking up speed. Only when the Angle of Attack is within normal operating range should the pilot start pulling the nose up - slowly, in order not to exceed the structural limit of 2.5g.
What happens in a stall in the Coffin Corner is described in this answer.

The higher an aircraft flies, the stall speed and critical mach get closer until at the absolute ceiling they are the same value.
Lets say you are in an airliner that is 5,000 feet below absolute ceiling and the two speeds are 40 knots apart and you are flying right in the middle of the two speeds. The pilot pulls the stick all the way back so that a full stall develops. At stall the nose drops. The weight of the aircraft is now accelerating downwards, possibly out of control, with engines still on cruise power and the critical mach is so close.
While this simple example would be very hard to imagine, due to stick shakers, fly by wire etc, there have been accidents in the past due to these reasons, where confusion or other factors have led to the initial loss of control.
I would have thought that a better pilot is needed on the recovery (to recover while preventing over speed) than the one that got you into the situation to begin with, so pilots are trained (and electronic aircraft designed) to avoid the situation in the first place!

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text:
Does throttle control altitude and yoke control airspeed?
I read that Throttle controls altitude and control wheel/ stick controls airspeed. Is that true? if so please explain.

you should read how it flies. engines add energy that is used for improving both kinetic and potential energy

I was taught this, but I was taught it as a specific technique for a specific phase of flight (approach). Not as an aerodynamic principle.


text:
Were there ever instrument approach procedures utilizing only Four Course Range Navigation?
Looking for information on 4 Course Range instrument approach procedures.  I am aware of their navigational use as we developed the airway system, but were they ever used for final approach segment navigation? Does anybody have access to such an approach plate?

Yes such approaches have existed.
As early as 1944 the North America Low Frequency Radio Range (aka Four-Course Range and Adcock Range) network was quite developed. There are several approach plates using this type of radio aid, e.g. this one from May 1944 for Meridian:

Source
The station is used for IAF and FAF. The approach is done using 334°/154° headings. None of them, nor the station are aligned with a runway, this looks like a circling maneuver.
Other plates:

Fort Chimo (Quebec)
Joliet (1957)

Some editions of the AAF Radio Facility Charts can be bought online, but the edition of June 1944 has been scanned by North Texas University:

On pages 38-39 we can see the overall coverage.
Other pages show details by area.

This is the page for Mississipi with Meridian, the airfield we discussed above, at the center:

We see that two of the four identified directions are not connecting airway segments. This is the ones used to approach the airfield where the station is located. They cannot be aligned with the runway because they play their role in the four-course range: There are four beams radiating over about 90° in azimuth, with small areas overlapping (about 4°). The middle axis of such areas is the actual path:

While the station location can be selected, and there is a certain possibility to have unequal quadrants, the adjustment is still limited. One or two directions are used for the airway(s), they determine the overall orientation of the other antenna system (angle bisector), and it's anyway difficult to create a small aperture (i.e. a very directional beam) at this frequency. This explain why the approach is a circling one.
Complete list of US approaches
At some point modern standardized plates were issued:

(Source)
The image below can be found on this site dedicated to simulation, which lists about 400 approaches in the US and Alaska.
More:

On Youtube: The Four Course Radio Range.
On Wikipedia.

Longshot: Does anyone know of a source for Visio or ??? templates for making plates.I want to make some Range-A approach plates.

@BobDenny: I would suggest to look at QGIS aeronautical applications. QGIS is an open source GIS, you can read more about QGIS by ICAO. As QGIS is able to read some ArcGIS data (used by ESRI), I guess you can get templates after some research.

Oh this is interesting, thank you! I will return to this when I have time. Right now my work has me crushed.

Oh! THis is really interesting. I bookmarked this and will look at it in the future. I used to have QGIS on my system but I removed it last year when its Python messed up the Python I have installed for other things. I hate the way one program can scorch another if they both use Python and then put s**t into the PATH.


text:
How do I select an appropriate controller gain for this roll control autopilot?
I'm trying to select a suitable K value for this roll control autopilot, based on:

Instead of the Gyro, I've also just considered a negative unity proportional gain feedback. How do I select the appropriate controller gain K for the system? (I'm feeding a step impulse into the system). I know that for each aircraft it would be different, for a fighter jet you want quick response times, but maybe not so much on a commercial airliner. What I'm trying to find out/understand is there design criteria/a method to selecting this value?

Is there a overshoot(%) I should aim for?
Is there a required damping ratio set?
Or do I eyeball it, and find a balance between the time to settle and overshoot(%)?

You want fast response, but not any overshoot. The perfect response to a step input in this case would be a critically damped one. Matlab/Simulink are your friends here.
Things to consider in your feedback loop:

Aircraft dynamics - is the initial inertia response modelled? Meaning, does the aircraft roll spool up to its final roll rate upon aileron deflection?
Is aerodynamic resistance to roll modelled?
Is it a digital loop, and if so what integrators are you using? If Euler your loop rates need to be high.

I've seen a similar design for a helicopter with a controller gain of 20 and an additional roll compensator $\frac {s + 1}{s}$ in the forward loop, and a negative unity feedback.
You could eyeball the response, use different settings for the feedback gain, and find the fastest response without overshoot. You'd have to testcase it for several extremes. You do need some form of damping feedback as well, I reckon your first task would be to find a response that is not oscillating.

<You could eyeball the response, use different settings for the feedback gain, and find the fastest response without overshoot.> There are analytic methods for that.

What sort of methods are you referring to? Plotting the root locus?

@Liam for example. then the Bode plot, or Nyquist. The peak of the overshoot is given through some simple formulas: https://en.wikipedia.org/wiki/Step_response#Control_of_overshoot http://ecee.colorado.edu/~ecen2270/materials/TransResp2ndOrdSys.pdf [eq. at page 8, unfortunately overlapped with the picture]

@Federico thanks, that's helped me clear up my understanding of what I'm trying to achieve here.


text:
How frequent are unpublished holding patterns outside general aviation?
Flight regulations say that ATC may put an aircraft in a holding pattern, which is not published in the charts but I couldn't find any such case in FlightAware.
I have been searching on the Internet but I think I found almost only some information about flying uncharted holding patterns in general aviation and that it is something the ATC would like to avoid.
I'd appreciate any information on how often (and when or where) these uncharted holding patterns are used outside general aviation.

Not very frequent at all.
The need to delay traffic toward an airport is quite common. Usually, it is simply a question of fitting someone in a sequence of arriving aircraft, but there can be more severe cases, such as a runway change or ongoing emergency. In any case, ATC has a number of tools to use, and unpublished holdings are usually one of the last tools to be selected.
What we usually prefer to do is make use of radar vectoring instead. This is much more flexible, easier to work with and requires less time on the frequency. A normal holding pattern takes 4 minutes to fly. If needing to delay a flight for just a few minutes, using a holding will only really work if it works to delay the flight 4, 8, 12 etc. minutes. If I need to delay someone 6 minutes to fit in an arrival gap, then I could put them in a holding for 4 minutes and then vector them around for 2 minutes, but if I am going to be giving delaying vectors anyway, skipping the holding is easier.
On a busy radio frequency, available time on the frequency is everything for a controller. Nonessential or long messages can really ruin your day when working a tight approach sequence into a busy airport. Let's look at the phraseology for instructing an aircraft to enter a non published holding that the crew were not expecting:

SAS123 proceed direct to ANDIE, alpha november delta india echo, and hold. Inbound track 120, right turns 1 minute legs.

Depending on your rate of speech, such an instruction may take almost 10 seconds to issue (and you can't just speak faster, because then the pilots will ask you to repeat yourself because they missed something).
If ATC needs to delay flight for a long time (say, more than 10 minutes), then holdings can definitely be an option, but in that case published holdings are used as much as possible. Published holdings are designed to ensure terrain clearance down to a specified lower limit, and, importantly from an ATC perspective, have defined lateral separation minima to other traffic. We are not allowed to use radar separation to aircraft established in a holding, so separating traffic from someone holding in a non published holding would be quite a hassle - but for published holdings, holding areas are defined, to which we can use radar separation. Finally, the phraseology for entering a published holding is much shorter:

SAS123 hold over ANDIE as published.

If ATC needs to delay a flight for a short time (say, less than 10 minutes), using radar vectors is usually much easier, both for the controller and the crew. It is much more flexible (I can vector people around, and start creating a sequence 50 miles from the airport, even though there is no room near the airport yet) and can be used to delay flights any number of minutes (not just multiples of 4). And the flight crew just needs to adjust their heading, no need to calculate holding tracks, compensating for current wind and so on.
Of course there are cases, for example emergencies, where pilots may request to hold somewhere that does not have a published hold, and ATC will of course permit them to do so. But looking at the workload establishing a non published holding gives us, using a published holding or some other tool (radar vectors) is usually the preferred option.


text:
Where is the toilet smoke detection indication displayed for a B777-200/300?
In case of smoke detected in any of aft toilet of a B777-200/300, where is the indication displayed? Is this at a single location, or at different locations?


text:
How are the navaids that are on the airport ground powered?
How are navaids (such as the ILS localizer) that are on the airport ground powered? Are they connected to the airport's power? Do they have their own backup power?


text:
How are Mode 2 pulses encoded?
I'm working on a research project looking at Mode 3A/C replies and have noticed some pulse patterns that look very similar which I suspect are Mode 2 replies.  They appear to differ by just one pulse.
I'd like to extend my decoder to decode these as well but I'm struggling to find out how Mode 2 pulses are encoded.  I know that mode 3A/C is encoded using the Gilham code but that excludes the extra pulse that I am seeing.
Is there a publically available document which explains how Mode 2 pulses are encoded ?  I've search high and low and haven't found one yet.

Let us continue this discussion in chat.

Mode 2 is classified, no?

Mode 2 isn't classified but the information is not easy to find.  After a lot of Googling yesterday I eventually found a document that provides me with the information that I needed.

I think the problem is that you see a Mode 3/A (identity) reply, but you try to interpret it as Mode C (altitude reply).
In Mode C, altitude is Gillham encoded, using all pulses except for D1 and X. This gives a range of -1200 to 126700 ft in 100 ft increments.
If you see a D1 pulse, you are not looking at a Mode C reply.
In Mode A, a 12 bit identity (squawk code) is transmitted, using all pulses except for X. In Mode A, the D1 pulse is commonly used (in fact, evey squawk code where the last digit is odd uses the D1 pulse).  It is simply the octal conversion of bits / pulses [A4 A2 A1], [B4 B2 B1], [C4 C2 C1], [D4, D2 D1].
For example, if you would encode squawk code 2345 you would set pulses [A2 = 2], [B1 & B2 = 3], [C4 = 4] and [D4 & D1 = 5].
Note that in case of Mode A, there may be an addition pulse for radar IDENTificaton. This is the Special Purpose Identification (SPI) pulse and it occurs 4.35 microseconds after the start of the closing framing pulse, with a duration of 0.45 microseconds.

source: www.radartutorials.eu


text:
How to extract the interrogator ID from a DF11 message?
I've written a class which decodes raw Mode S messages.  Whilst browsing the ICAO specs the other day I noticed that the Interrogator ID is sent back in the DF11 message.
I thought that it would be a fun exercise to determine where in my coverage area interrogations from particular radar installations can be seen.
But I'm having a really hard time figuring out how to extract the interrogator ID.
For this DF11 message:
5D406FDB0D62DD
I've got the following:
Binary: 01011101010000000110111111011011000011010110001011011101
Integer Words:
    [0] => 93
    [1] => 64
    [2] => 111
    [3] => 219
    [4] => 13
    [5] => 98
    [6] => 221

 CA: 5
 AA: 4222939
 Address: 406FDB
 Parity: 255
 Residual: 0

As the Residual value is 0 I know that this is a valid DF11 message and the address checks out as well.
From the ICAO spec:
"The code used in downlink PI field generation shall be formed by a sequence of 24 bits (a1, a2,..., a24), where the first 17 bits are ZEROs, the next three bits are a replica of the code label (CL) field (3.1.2.5.2.1.3) and the last four bits are a replica of the interrogator code (IC) field (3.1.2.5.2.1.2)."
In that case I would expect the end of the last three integer words (4,5 and 6) to be mostly 0's but they aren't.
I've gone through the documentation numerous times but just can't see how to get the II and SL codes out of Parity value ?
I'm calculating the Parity value by doing the following:
$this->_pi = $this->_int[4] | $this->_int[5] | $this->_int[6];

where | in PHP is the OR operator.  Is this my mistake ?

Your calculation of $this->_pi just ORs the three parity bytes together,
which certainly doesn't make any sense.  I'm not sure what you hope to get out of that. Even XORing them together wouldn't seem to serve any purpose.

Addressing and checksumming in the Mode S protocol work in a funky way, in order to save bits. The last 3 bytes of a message are constructed as a CRC checksum (which Annex 10 calls "parity") of the first 4/11 bytes, XORed with an address for the "conversation" the message is part of. When you receive a message you're supposed to calculate the raw CRC on your own and then XOR it out of the bytes. What is left is the address, and by comparing it to addresses you're interested in you can see if the message was for you, or alternatively either garbled or for someone else.
In the basic Mode S interrogation protocol, this "address" is just the airframe address, and the checksum-xor-address field is named AP.  For uplink messages, transponders know what their own address are, of course -- and for downlink, radars know which addresses they have interrogated recently.
In a few other cases, the "address" is the "interrogator identifier", and the checksum-xor-address field is named PI.
For DF11, the transponder knows which II to put into its replies because it gets them out of the all-call message, and the radar knows which II corresponds to its own all-calls, so it can recognize those responses.
Finally, in squitter where the transponder sends a downlink message without being prompted by an uplink message, it sets the "address" to 0.

If you have been decoding ADS-B broadcasts, you're listening for squitter -- so you can identify them as the ones where the CRC you compute equals the PI field in the message (because XORing 0 into the CRC doesn't change it). Everything with a different address will look like checksum failures to your code.
Since the DF11 message you're showing does match the cheksum, it must be DF11 squitter. There are plenty of those to hear; TCAS depends on everyone sending them continually.
In order to find DF11 messages with a nonzero interrogator identifier you should look for messages where the last three bytes after you've XORed out the CRC do begin with 17 zeros but then have at least some of the remaining 7 bits nonzero.

Thank you for the very clear explanation :-)  I'll revisit my code tomorrow and will post back but I can see exactly how it should work now.


text:
What is the difference between a shock wave and an expansion wave?
Studying high speed aerodynamics I have found a question asking for what happens to static temperature for an airflow passing through an expansion wave.
The correct answer states that it decreases- the opposite of that on a shock wave.
What is the difference between these two waves that makes them contraries?

An adiabatic expansion lowers temperature, an adiabatic compression raises temperature, analogous to the way a refrigerator and an air conditioner work. Or a bicycle pump, it gets quite hot while you're pumping air.
A shock wave compresses air - at supersonic speeds the object travels faster than the pressure information, and when the object arrives at a bit of air it compresses suddenly in a shock wave, with corresponding higher temperature.


text:
How does a fixed-pitch propeller changes the blade's angle of attack?
Studying I have found a question that states that in a fixed-pitch propeller in flight at a given TAS, the blade's angle of attack will change if RPM increase, but I don't understand how can the blade's AOA change if the pitch of the blade is fixed.
Am I missunderstanding the concept of fixed-pitch propeller or the question is wrong?
Question answer and explanation for source:
Q:For a fixed-pitch propeller in flight at a given TAS, the blade angle of attack will:
A:increase if RPM increases.
E:For a fixed pitch propeller the pitch angle cannot be changed (it varies from root to tip to maintain near-constant angle of attack along the blade).
The angle of attack as in the case of a aircraft wing is defined as the angle between the chord line and relative airflow. With a propeller however the relative airflow is the resultant of the airflow due to rotation and forward speed. Change either of these values and the angle of attack will change.

related: https://aviation.stackexchange.com/questions/33481/how-does-the-relative-wind-on-a-propeller-strike-the-blade-from-behind-the-non

Okay nice, it's related but it does not answer my question. Thank you though.

The pitch of the blade is fixed.  But the angle of attack depends on how the blade moves through the air.  That motion is mostly a combination of the forward speed of the airplane and the rotational speed of the blade.
In your graphic, the blade is attached to a plane that is flying up the page.  The blade is sticking out of the page and is being pushed to the right by the turning engine.
At some combination of airplane speed (up the page) and propeller speed to the right, the air would flow exactly along the (fixed) pitch of the blade.
Now, without immediately changing the speed of the plane you increase the RPM.  The blade moves to the right more quickly.  The angle of attack it makes in the air increases.
Put your hand out a car window.  Hold it at a fixed angle relative to the ground (say 15 degrees).  Now, holding the angle constant, move your hand downward rapidly.  The pitch has not changed, but the angle of attack has.  This is the equivalent of increasing RPM.

The example of the hand out the car window really made me understand it. Just great.


text:
Why did my transatlantic flight meander off the great circle?
I recently flew commercial to Scotland and was tracking our progress on my iPad using Foreflight. I made a direct-to line that followed the great circle route, but noticed that we meandered off of it quite a bit.
The only explanation I can think of is that the crew was trying to take advantage of favorable tailwinds by following the jet stream. Is this assumption correct?
And if so, can an autopilot be programmed to follow a route dictated by a winds aloft forecast?

Most flights across the Atlantic don't fly direct great circle routes; they use the North Atlantic Track system, which is an array of "lanes" you might say, usually 60Nm apart, like one of those slot car racing places with 6 or 8 lanes of cars.  Just about everybody that can make the trip non-stop will be in the tracks at various altitudes. Off each coast are an array of fixed entry point fixes called Oceanic Entry Points as jumping on or jumping off points.
The north/south position and the precise routing of the tracks is set up each day, with the focus on "a westbound flow departing Europe in the morning, and an eastbound flow departing North America in the evening." (ICAO Doc 007) Tracks are moved up and down the take advantage of winds or weather, especially jet streams.
Jet streams flow along air mass boundaries at the top of the troposphere and change constantly as air masses move.  The tracks are set up to try to exploit or avoid them depending on whether they are favourable or not (as well as other weather systems).

can an autopilot be programmed to follow a route dictated by a winds aloft forecast?

Yes, but the question is how it is done, and these days it's better to think of it as not just the autopilot, but a Flight Management System (FMS).
In the case of the North Atlantic Track system explained by John K, the waypoints that define the route an airplane is to take are defined by latitude and longitude.
In my day (old fart-retired in 1999) the waypoint lat/long values were read to us over the radio, and we entered them into inertial navigation system system (actually 3 of them) which then controlled the autopilot—they voted so to speak.
These days there are a better ways of getting the assigned route into the FMS, and which hopefully someone familiar with them can explain.
The point is, though, that the FMS is not looking at the winds aloft forecast and deciding the route, it's simply flying a route given to it by some means that has been ordered by a ground controlling facility to take advantage of the winds.
Now, could an FMS be programmed to receive the winds aloft forecast and decide from that the best route. Probably so, but when you add in the complications of separation for all the other aircraft doing the same thing, I don't think we're there yet. In other words, automated ATC is still down the road a ways to say the least.

The North Atlantic (NAT-HLA) High Level Airspace is special use airspace where a series of optimized routes called the OTS, Organized Track System, are published every day, twice a day. This airspace is unique for having a largely unidirectional flow of traffic, and the waypoints to be followed over the Ocean are not fixed. Your assumption "following the jet stream" is part of the answer and the following 2 extracts from ICAO NAT-HLA Doc. 007 should effectively serve as a more complete answer to that question.

2.1.1 As a result of passenger demand, time zone differences and airport noise restrictions, much of the North Atlantic (NAT) air traffic contributes to two major alternating flows: a westbound flow departing Europe in the morning, and an eastbound flow departing North America in the evening [...]

2.2.1 The appropriate OACC constructs the OTS after determination of basic minimum time tracks; with due consideration of airlines' preferred routes and taking into account airspace restrictions such as danger areas and military airspace reservations. The night-time OTS is produced by Gander OACC and the day-time OTS by Shanwick OACC (Prestwick) [...]

The routes actually flown are created by the company Flight Dispatcher and provided to flight crew as an "Operational Flight Plan" or OFP in paper or digital format. This plan is sometimes called a Fuel plan because it determines the legal minimum fuel required. Most of the time it represents the most fuel efficient routing but it can also be set up for time efficient routing or routing manually strung up by the Dispatcher etc. Overflight charges (fees) are also factored into the cost optimization. For longer flights especially, a 'route search' is carried out as there maybe many route options available and an effective choice can result in much cost saving.
The Flight Planned route and other details are fed to the on board FMC (Flight management Computer) either manually by the pilots in case of a paper flight plan or, the plan can be up-linked by the Dispatch office using electronic means, and checked, accepted and activated by the pilots in the cockpit with a few button presses. Capability exists today for the pilots to accept route change suggestions by Flight Dispatch whilst enroute in real-time thanks to realtime monitoring of the flight progress and weather/wind conditions and subject to ATC clearance.
At the current state-of-the art, and with increasingly reliable datalink capabilities, the grunt work of optimal flight planning is left to a specialist Flight Dispatcher and sophisticated flight planning software. Comparatively, the on board FMC can do rudimentary optimization with respect to winds aloft as it only knows the limited wind forecast fed in by the pilot or uplinked from Flight Dispatch. The Autopilot/FMC combo can fly a given route, but it cannot create that route.
And Yes, the auto-flight system's normal mode is to couple the autopilot to the flight plan loaded into the FMC, infact this is the way most of the navigation is done in today's airplanes.
Finally, some points on GC flying -

'minimum distance' (=Great Circle) is one of many factors upon which flight planning is dependent.
The internal FMC navigation functionality is such that it strings and flies each segment between waypoints as a Great Circle.
If ATC was able to allow free, 'point to point' routing from the departure direct to destination, it would be one GC track all the way.
What also hinders us, is the fact that we have to follow Airways, and traditionally, Airways were from radio beacon to radio beacon, and beacons were located at airfields and towns/cities and these were rarely coincident with the GC.
Flights over the oceans and other remote areas may have a better chance of conforming more closely to the corresponding GC.
ATC already clears us for "direct" routings at the tactical level, in real time. Their concern for a green world is the same as our's!
Despite being an oceanic area, NAT flights are more rigid about maintaining the planned (non-GC) route and they aren't able to agree to many direct routing requests because of the nature of the operation, in turn dictated by the sheer volume of traffic and it's mostly unidirectional flow.
Todays Flight plans can even give us niceties like a Great Circle Index or GCI which tells us in percentage form how much more distance we are flying compared to the direct GC (though this info may not be of much practical use in light of the more sophisticated optimization  mentioned above).


text:
What modifications are required for automotive fuel STC?
What actual engine modifications are required to allow certificated piston engined general aviation aircraft to run automotive fuel under an FAA Supplemental Type Certificate (STC)?

Which STC?  They are available from Petersen and EAA.  Or are you asking what would it take to get your own STC?

I am asking about the physical modifications to the engine that are required to run automotive fuel.

It depends on the particular airplane, I think.  When we did a autogas STC on our Cherokee 180 (some years ago now), there were no modifications to the actual engine, but we needed to replace the OEM fuel pump with a dual-pump setup, and replace some of the fuel lines.

@jamesqf, thanks for that.  Do you recall if these particular modifications were specifically required to comply with the STC, or were they incidental?  In otherwords, while inspecting the fuel system for the STC might you have noticed that some lines were old and just decided it would be a good idea to replace them while you were at it?

@Michael Hall: As I recall, everything that was installed or replaced was required, and came in a kit with the STC - pretty sure it was Petersen: https://www.autofuelstc.com/piper_airplanes.phtml

Every O-320 and O-470 equipped plane I have flown using the STC required no modification whatsoever.  High wing or low wing, single or twin.  Some of the larger engines may require water injection but I've never actually seen one in the wild.

For most engines it would require a hardened valve seat.
The reason why lead is required for aviation engines is that the value seats are not hardened and the lead in the gas helps prevent "micro-welding" that can occur on a valve seat.  Micro-welding happens when the valves are hot and the the valve is pressed against the valve seat; the valve will start to weld to the seat before it is torn away when the valve opens.  This causes a jagged edge at the valve seat which causes a bad valve seal.  Old engines relied on the lead in the gas to prevent this. When the automotive industry switched to unleaded gas, the engine makers had to use valve seats out of hardened materials that were more resistant against heat.
The lead is also a great octane booster although many airplane engines can use octanes lower than 100.  In automotive fuels today ethanol is used to boost octane and that has its own host of problems since ethanol is very corrosive.  So before you switch to automotive fuels you need to make sure you engine and fuel system can handle the ethanol.
For a discussion of micro welding among mechanics check this link out.
EDIT:  After doing more research I discovered that Lycoming has always used hardened valve seats so the modifications on a Lycoming engine would be very minimal.  However, most Continental engines do not have hardened seats and major modifications would be required.
Lycoming has a good article on the use of unleaded fuel in the their engines here.  Notice however that Lycoming states that ethanol CANNOT be used in their engines.  In the US at least it is getting very hard to find fuel that does not contain ethanol.

Thanks DLH, I figured any changes would be related to lead, octane, or maybe both, but didn't consider ethanol.


text:
What are the WAAS requirements for RNAV (GPS) approaches?
Is a WAAS GPS required to fly in RNP airspace (e.g., RNAV (GPS) approaches)?
Or, can you be approved to fly RNAV (GPS) approaches with, say a military EGI (embedded GPS/INS)?

RNP enroute is different from an RNP approach, which is different from an RNAV approach.  Also, which minimums lines are you inquiring about? LNAV? LNAV/VNAV? LPV? LP?


text:
Where in the FAR or AIM is a high speed taxi defined?
While monitoring coms I recently heard a Cessna Centurion that had just cleared the runway request a "high speed taxi", which was approved by the tower. I'm assuming that means he wants to taxi at a faster than normal rate to his FBO of choice. Are high speed taxis defined somewhere as something you can request of ground control?

Closely related

Yeah @Pondlife I saw that - it is definitely related, but my specific question is whether or not there is something in the AIM or FAR that backs the radio coms I heard. I'd never heard another pilot request "high speed taxi" -- but it's not like i've logged 10,000 hours so that might just be my inexperience :)

There's no definition in the AIM or regulations, as far as I know, but there is one in the P/CG, sort of (emphasis mine):

HIGH SPEED TAXIWAY− A long radius taxiway designed and provided with
  lighting or marking to define the path of aircraft, traveling at high
  speed (up to 60 knots), from the runway center to a point on the
  center of a taxiway. Also referred to as long radius exit or turn-off
  taxiway. The high speed taxiway is designed to expedite aircraft
  turning off the runway after landing, thus reducing runway occupancy
  time

So at least as far as the FAA is concerned, a high-speed taxi may be up to 60kts.
As for requesting it, you can request more or less anything. There's an active EAA chapter at my local airport and pilots who are in the process of building and testing aircraft sometimes request a high-speed taxi on the runway. The controllers don't question it or set any speed restrictions, it's all on the pilot. I suppose if things were busy they might deny the request, but the pilots know to do these things at quiet times anyway.

Remembering that for some aircraft taxying at 60kts is called 'taking off'.

A high speed taxiway (called rapid exit taxiway elsewhere) is specifically for vacating a runway. It doesn't seem to apply in this case.

He was likely asking for approval to use the runway for a simulated take off without actually departing. Usually testing engine performance after modifications.

I think you're right. The parking he was heading towards is at the end of KTOA. He probably landed and was asking if it'd be ok to tie up the runway to get to the end of the airport faster.

This is the correct answer. Use case here is: after getting major repairs you want to test the plane in a series of tests. Short run up, long run up, taxi, high-speed taxi (enough to take off but dont), 1 lap around the pattern, couple laps, circle the airport for a while, then go out for a flight.

The FAA does not provide a taxi speed limit but most airlines require observance of SOP "standard operating procedures" as part of their certification process. Normal SOP's are 30kts max with 10kts preferred. The rule I have seen for most flight training schools is "no more than a fast walk".
While not specific to the FAA's FAR or AIM, this FAA video describes the design and requirements of high speed taxies and refers technical specifications to AC 150-5300-13A-chg1. They define a high speed taxi as having no more than 30deg and nor more turn radius than 1500ft.
The exact dimension depends on aircraft category, IFR minimums, whether a reverse turn is needed, etc. No speed requirements are required, they define the accommodation for speed as "higher than normal taxi speed".

Taking a "High Speed Taxi" refers to using a specific taxiway designed to exit the runway at a high speed after landing. The term does NOT refer to using ANY taxiway at a high speed. Larger busier class D, C, and B airports have these highspeed taxiways.
An example of this is Taxiway Hotel at Van Nuys off of 16R. Thus the pilot you mentioned was requesting "the highspeed" taxiway to exit after landing. Whether he did that at a slow rate or fast rate was up to him. I frequently use the above taxiway at VNY because instead of taking the other taxiways at a 90 degree angle (which requires slowing down to almost a crawl to exit the runway), the highspeed is at a 30 degree angle to the runway which allows aircraft to exit the runway much faster than conventional taxiways. Busier airports want to get planes on and off the runway as fast as safely possible to help the flow of traffic. Highspeed taxiways help this process.

Welcome to Av.SE

Welcome to ASE, but wouldn't that airport feature be referred to as a "High Speed Taxiway", as noted in another answer, rather than a "High Speed Taxi"?


text:
How does the range of a single- and twin-engine for small aircraft compare?
I know that one difference between these two types of small aircraft are that in a twin engine, one engine can continue working if the other suddenly fails.
Wouldn’t a single engine plane weigh less and thus can fly longer on a specific amount of fuel comparative to a twin engine plane?

Possible duplicate of What are the pros and cons of single-engine vs. twin-engine?

Answers to the proposed duplicate don't mention range as pro or con. Not a duplicate.

If you ask it this way, it will be hard to answer because there is hardly an existing pair of instances with everything else being equal. However, you can refactor the question to like "if we design an airplane to be twin engine instead of single engine with all the same performance, would fuel economy be worse?" Or "is there any gain in fuel economy to minimize number of engines" that will be more abstract and easier to answer.


text:
Are there specific names for airport vehicles used for de-icing?
Do de-icing vehicles used at airports for de-/anti-icing procedures have special names? I-ve searched the internet but have not found the exact names of these vehicles.

As far as I can tell they don't have their own name specifically they are part of the airports Ground Support Equipment. They have non-powered and powered support equipment, De/anti-icing vehicles fall under the powered category. Trying to find a better link than the one I found on Wikipedia In case you want to see that link
In case you want to buy one Heres a manufacturers website of GSE


text:
What is the relationship between the terms "airprox" and "near mid-air collision"?
Do the terms "Airprox" and "Near mid-air collision" mean the same? If not what is the difference between them?

Where did you see these terms?

AIRPROX is the phraseology recommended by ICAO, midair collision, close call and near miss can also be used, but not as a standard phraseology!

So under ICAO recommendations, if you collide with another plane and it's somehow not a mayday scenario, you don't say anything other than that you came close?


text:
Can a lightning strike lead to structural damages of the fuselage and eventually cause decompression?
I know that modern aircraft are designed to withstand lightning strikes. Still is there a possibility for aircraft to sustain significant damages of the fuselage due to lightning strikes and subsequently face decompression?

There's a possibility that a meteor will whiz through the cockpit and take out the pilots... I'm not sure what you are really asking? Is it possible? Yes, but probably close enough to be statistically zero...

What he is asking is a legit question.  Lightening strikes happen all the time, just not on flat surfaces like the pressure hull. Usually somewhere on the wings, nose or tail.   Composite sections of airplanes like fairlings have wire mesh laminated in under the surface to be conductive for lightening strikes and for electrical bonding purposes.

Decompression is a tricky thing.  The aircraft is already quite "leaky" and without a steady flow of pressurized air from the engine bleed air (or compressors, in the case of the 787) the altitude in the cabin will rise slowly. There is a pressure controlled outflow valve in the back of the aircraft that is usually about 10-20cm square. The pressurization system stuffs air in constantly, and the outflow valve opens and closes to keep the interior pressure at the correct altitude.  Any hole smaller than the amount of air the pressurization system can pump in will just be handled by the outflow valve closing somewhat.  Depressurization due to structural failure is fairly rare.  Most recent would be Southwest 1380, where an uncontained engine failure broke a window.
I have not heard of a lightning strike even piercing the aircraft fuselage in modern civilian aircraft, much less in the cabin area.  There are a few cases I have heard of it happening to military aircraft, and in those the hole was fairly small (about 2-5cm in diameter).  Aircraft are regularly hit by lightning, and extensive testing is done to simulate lightning strikes before an aircraft can be certified.

Compressors, in the case of the 787... or the 707.


text:
What's the difference between SLG strut used in B777 and truck position actuator used in any other aircraft like B747?
Do they serve the same purpose ? If not what's different ?

The SLG strut (found on B773ER) is a truck position actuator on steroids.
The truck position actuator is used to bring the landing gear truck into a position suitable for stowing in the wheel well. This is true for all Boeing aircraft.
The SLG (Semi Levered Gear) strut on the B777-300ER is additionally used to increase T/O performance. During T/O the wheel truck is locked perpendicular to the strut, shifting the point of rotation from the middle of the truck to the aft wheel. Thereby the rotation pitch can be higher and the tail clearance during rotation is increased by about 30cm / 12in.
After selecting gear up it behaves like a normal truck position actuator, moving the landing gear truck into stow position before the gear is retracted. During landing it is disconnected and serves only as a shock damper.
Source:

FCOM B777-300ER
http://boeing.mediaroom.com/2003-06-18-Semi-Levered-Gear-Performs-During-Boeing-777-300ER-Flight-Tests


text:
What's the difference between Screwjack and Ball screw?
Flaps are driven by screw jacks or ball screw ?

The text of your question and the posted question ask two different questions.  Are you really asking about both?

Can you specify what you mean

A ballscrew is one fancy type of screw you could use in making a screw jack, so the question is a bit meaningless, not to mention far too broad.

A ball screw is a type of jack screw. I'm guessing you are looking for the difference between a Ball Screw and an ACME Screw. Where each are used is dependent on the aircraft. Boeing designed jets all use Ball Screws everywhere. MD liked to use ACME screws.

A ball screw uses round bearing balls inbetween the screw shaft and the sliding nut it engages to transfer the forces between them, with a minimum of friction. The spiral grooves cut into the shaft and the nut are contoured to accomodate and hold in place the balls between them. A screw jack has the threads cut into the rotating shaft in direct contact with the threads cut into the slider nut. The ball screw mechanism is more precise, resistant to wear, and easier to move but more expensive than the screw jack.

All true, but which system is actually utilized for flap deployment?  And on which type of aircraft?  While the answer evaluates different methods of flap actuation, it doesn't answer the text of the question.

I answered the question posed in the title. My answer is an accurate answer to that question and does not deserve a downvote.

Agreed -- you answered the question in the title, however, the question asked in the post is unique from the question in the title.

@Marius - the question in the body of the question is obviously too broad, as it is inconceivable that this has not varied between aircraft.  Probably it should be closed.

Or perhaps slimmed down to a single kind of aircraft, as an example.  That said, if the OP isn't going to come back and do something about it, I'd move for closure.
Also, for reference, I removed my downvote after the first couple comments.

this question has generated a surprising amount of interest (as measured, perhaps inaccurately, by the score that the answer has accrued), even if the OP's statement of it may have been too broad or unclear.

Don't forget that ACME screws are self-locking. Where each is used depends on the aircraft and more generally the OEM. Boeing doesn't use ACME screws, while MD loved them.


text:
What is the definition of Lift?
Lift is a term that is used very frequently here, but I cannot find any post explaining what it is exactly. I am looking for the exact definition of lift, not an explanation of how it is generated.
It seems a very simple question, but I have not been able to find consistent answers to it. Some very well respected sources even contradict each other:
At https://en.wikipedia.org/wiki/Lift_(force) Lift is defined as:
A fluid flowing past the surface of a body exerts a force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction.
But according to NASA (https://www.grc.nasa.gov/www/k-12/airplane/lift1.html):
  Lift is the force that directly opposes the weight of an airplane and holds the airplane in the air.
If a plane is flying straight and level (at constant altitude) these definitions are the same, but otherwise they are not the same at all. So, which one is correct?

One is a technical definition, the other (NASA site) is a layperson's description for the general public.

This question may be helpful

It's what keeps you from going smashy-smashy.

The quote from NASA is only the introductionary first sentence. The rest of the paragraph provides the actual definition: "Lift is a mechanical aerodynamic force produced by the motion of the airplane through the air. Because lift is a force, it is a vector quantity, having both a magnitude and a direction associated with it. Lift acts through the center of pressure of the object and is directed perpendicular to the flow direction. There are several factors which affect the magnitude of lift."

@Koyovis That is true, the word perpendicular is even printed bold, even though it does contradict the first sentence. I did find definitions like the one in the first sentence in several other sites too:  https://web.mit.edu/16.00/www/aec/flight.html, "In order for an aircraft to rise into the air, a force must be created that equals or exceeds the force of gravity. This force is called lift." and https://www.britannica.com/science/lift-physics "Lift, upward-acting force on an aircraft wing or airfoil." I figured this could be confusing to many enthusiasts so I asked this question.

It's the component of the aerodynamic force perpendicular to the direction of the air flow, the aerodynamic force being the reaction of the wing to the relative wind.

What if we introduce a bank angle, how would that change the direction of the lift, would it rotate with the plane or stay upward? And if we now keep this bank angle constant and also add a yaw angle, would the sideway force due to the yaw angle be included in the lift?

@Orbit Pure bank has no effect on aerodynamics. The actual definition of lift under yaw depends on whether you're defining it in stability axis or wind axis. Most define it under the stability axis.

@JZYL Bank does not influence aerodynamics, but lift does rotate with respect to gravity, which is interesting for a complete understanding. As xxavier states, lift is defined relative to the incoming flow, but even when decomposed to the aircraft coordinate system, the direction changes when yawing.

I'd fine tune the definition slightly.... Molecules of air hit every point on the airframe, creating aerodynamic force.  Lift, then, is "The vector sum of all aerodynamic forces on the aircraft that lie normal (perpendicular) to the flight path of the aircraft through the air." Regardless of whether that vector point up forwards, backwards, sideways or upside down. Just as drag is the component that is parallel to the flight path.


text:
Is aspect ratio efficiency related to the momentum added to the air?
Many years ago I was taught that for generating lift or propulsion, lower losses were associated with imparting less momentum change to a larger volume of air, and higher losses were seen when imparting larger momentum changes to a smaller volume of air - and that the losses were due to friction (i.e., the air's viscosity).
The propulsive example of this is the commonly-cited turbojet-versus-high bypass turbofan engine comparison, and I had been under the impression that the airfoil example was high aspect ratio versus low aspect ratio wing designs. However, having just learned that this is in fact incorrect (it has to do with induced drag instead), is there in fact any connection at all between the imparted momentum change/volume of air argument and airfoil aspect ratio?

You're almost there. The difference lies in the kinetic energy required to move a certain mass of air, not in the air's viscosity. Copying freely from one of my earlier answers:

The force to keep the object aloft is $$F=m_{object}g$$ The force generated by downwards momentum transfer is $$F=\dot{m}v$$ with $\dot{m}$ indicating mass flow (kilogram per second) of the air (not the mass of the object). The energy flow (power) required to  impart this momentum on the airflow is $$ P=\frac{1}{2}\dot{m}v^2$$ Here we can draw an important conclusion. The power requirement is arbitrarily small, by increasing the mass flow and decreasing the downwards velocity.

The power requirement expresses itself in the form of induced drag. By affecting a large mass of air (a long wingspan), you reduce the power requirement, which manifests itself in the form of lower induced drag.

this is most excellent, thanks for the clear explanation.


text:
What are the design principles that result in thin, long-span, narrow chord main rotor blades?
Regardless of the overall design configuration of a helicopter, it appears that helicopters have main rotor blades that all fall into the class of ultrahigh aspect-ratio wing design: that is, you never see a modern helo swinging blades around that look like the wings of a piper cub, for example. What are the design principles that result in thin, long-span, narrow chord main rotor blades rotating relatively quickly for every helicopter I can remember seeing?

Check out the Boeing X50 Dragonfly.

Unlike with fixed-wing aircraft, there are very few structural limitations on the planform of helicopter blades.

Pic above was used in this answer and shows the centrifugal force acting on the blades. The blade is hinged to the rotor hub, and from the angle relative to the rotor hub we can see that the lift force on the blade in vertical direction is much lower than $F_C$. That allows the helicopter to use blades with very high aspect ratio, which is aerodynamically the most efficient for a given lift area.
A rotor like in the pic has zero bending moment at the root: it has a hinge. Fixed hub blades do experience a root bending moment but this is also comfortably evened out by the centrifugal forces.
In general, the design principles for a helicopter rotor are:

For a given design gross weight & range, establish a practical rotor disk loading. This results in the rotor disk area => blade length.
Rotor rotational speed is as high as possible, without entering critical Mach on the forward sweeping blade in cruise.
Number of blades is the minimum we can get away with - rotors with the lowest solidity have the highest aerodynamic efficiency. Increasing the number of blades is sometimes necessary for lower rotor induced vibration and/or a less distinctive noise signature.

Regarding the last point: the Wop Wop of the UH-1 was a well known distinctive feature. The Huey had two blades with a larger chord than usual: due to the teetering hub design the number of blades was limited to 2, and in order to accommodate the required disk loading the rotor blades have a wider chord than on similarly sized fully articulated rotors.

What is the relationship between teetering limits and blade length that limits blade length?

@JohnK Was distracted, have amended, thx.

A shame that you didn't get an upvote when the answer was accepted. +1.

I would remove the statement "Unlike with fixed-wing aircraft, there are very few structural limitations on the planform of helicopter blades." Both have structural requirements, but most of those can be addressed thanks to the wide availability of a variety of materials. The limitations on fixed wing and helicopter blade planforms are governed more by design requirements, rather than structural integrity. There are many examples of high AR aircraft available that show that structural limitations can be overcome for fixed wing high AR aircraft.

@MishaP The statement is exactly the reason why. Wing /blade load and structural integrity iare the primary design criteria.

@Koyovis I agree, which is why I think "Unlike with fixed wing aircraft" is out of place. Not that it makes your answer worse.

The ratio of the blade area to the total swept disk area is called the solidity, usually expressed as $\sigma$.
The rotor thrust coefficient is $C_T$.
The average blade lift coefficient is:
$C_L=6\,C_T/\sigma$
For highest efficiency, you end up wanting as large a disk area as possible.  Lowest disk loading.
Typical values of $C_T/\sigma$ are 0.1 to 0.15.  This is called the blade loading.  The figure of merit (hover efficiency) is a function of blade loading.  It is essentially like trying to operate an airfoil at its lift coefficient for best l/d.
Traditional propeller and helicopter design typically drives to highly loaded blades with near sonic tip speeds.
However, the new push for ultra low noise eVTOL aircraft drives rotors in a different direction.  They're driving tip speeds very low and pushing $C_T/\sigma$ down (lower blade loading).  All this is in the name of noise.
The result of some studies is some pretty comical blades -- think 16' diameter and 3' chord.  Like swinging two doors around.
So far Joby and Archer (and other credible eVTOL concepts) haven't gone quite this extreme, but low noise pushes in this direction.

comment under question: "check out the ... X 50 Dragonfly"

Well, both crashed.  Reason: rotor created too much turbulence around the fuselage, creating an uncontrollable pitch up tendency.
In addition to @Koyovis answer, a very high aspect ratio (allowed by the "centrifugal force" on the blades as they turn) reduces turbulence, thereby reduces interference with the following blade (and anything else that happens to be around).
The reduction of turbulence per unit rotor lift produced is by the same principle of reducing drag in high aspect wings:  smaller, weaker trailing edge and tip vortices. This is also why high aspect propellers, as few as possible, are most efficient.

At least for subsonic flight and a fixed vehicle weight, adding chord to a wing basically lowers its stall speed and reduces efficiency at high speed.
So, airplanes that don't have to worry about stall speed do have thin blade like wings- gliders because they have long wingspans, and cruise missiles because they don't land. To fly slow with limited wingspan, airplanes add lots of chord, sometimes with slats, sometimes with delta wings, sometimes by just spamming wings- biplanes and triplanes.
Whereas airplanes stall at low speed, helicopters stall at high speed due to the retreating blade. So, slow helicopters like news helicopters don't have to worry about stall and have pencil thin wings, while fast helicopters have lots of chord- occasionally by adding chord to the blades like the X-wing, but more typically by spamming more blades.


text:
Can the energy dissipated in a power-off descent be calculated and expressed in horsepower?
If the weight in pounds of an aircraft is known along with its sink rate in feet per minute at best power-off glide conditions, can the horsepower being dissipated by drag as the plane descends via gravity be calculated?

Horsepower = weight(lbs) * sink_rate(fpm)/(550*60)
It really is that simple.
Background
At a constant airspeed glide, kinetic energy isn't changing, only potential energy. So we know that all the energy going into drag is coming from the descent.
A basic law of physics is:
$P = F*v$
where

$P$: the power
$F$: the force, in our case the force due to gravity
$v$: the velocity, in our case the sink rate

The only trick is that we have to make sure that units align. In the case of horse-power, which is 550ft-lbs/s, we have to convert fpm to fps (divide by 60) and then scale by 550 to get the answer in HP.
Even better is when we don't use non-standard units. Then there is no conversion factor, because power in Watts is identically Nm/s.

Most excellent! I had suspected that but wasn't sure whether or not I needed to first convert pounds to slug-furlongs per imperial fortnight, or something equally psychedelic. -NN

by the way, the sink rate has to be expressed in feet per second. -NN

@nielsnielsen slug-furlongs per imperial fortnight could be called "nielsens"? :)

I am honored! -NN

1 horsepower=550 ft-lbs/sec, not per minute, per Google supporting niels nielsen's comment

@RossMillikan aaaaand that, ladies and gentlemen, is why we must only use metric. :D

What? And lose the BTU?   Seriously, would the math be the same for water impounded behind a dam, e.g. if the airplane's weight of water fell 500’ through the dam's turbogenerator, is that the same energy as the airplane descending 500’?

@Harper-ReinstateMonica, please, lose all of it! Yes, the energy change is identical. Potential energy is simply $mgh$, where $m$ is the mass, $g$ is gravity, and $h$ is the height drop.

Then there needs to be a Merkel Thermal Unit or something I can say without sounding like I'm talking about something else. Because so many hard thermal problems are made easy to communicate by the BTU.

That power is sink rate x glider weight. If you use m/s for sink rate and newton for weight, you'll get the power in watt. Advantages of SI...
It's easy to understand why it is so. Within a uniform gravitational field with acceleration g, a mass m is pulled down with a force m·g. The potential gravitational energy E at a given height h is E = m·g·h
Now, for an infinitesimal variation of height dh, the corresponding infinitesimal variation of energy is dE = m·g·dh
And if that variation dE takes place in an infinitesimal time dt, the power implied is W = dE/dt = m·g·dh/dt. Since, in our case, dh/dt is the sink speed w, we have that power W = m·g·w

Not an advantage in this case, because you've got to convert the sink rate shown on your variometer to meters/sec, and the weight of your glider to kg.

Unless you know both of these things already.  BTW weight is in Newtons, not kg.

@Frog: Maybe for the pedantic, but I assure you that if you visit Europe (or other metricified) place, and buy things that are sold by weight, the package is almost certainly going to be labeled in kg.  I'm also willing to bet that if you look up the weight of European-built airplanes, it will be given in kg.

@jamesqf It's true, and may be due to the fact that the kilogram was, in the original definition, the unit of weight. In fact, and until the 1960s (more or less) many calculations were done with the MKS system, where the reference unit was the force, its unit being the kilogram-force or kilopond, (weight of the standard kilogram) and the mass was a derived unit, often called UTM (Unidad Técnica de Masa) or TME (Technische Masseneinheit), the amount of mass accelerated at 1m/s2 by 1 kilogram-force...

@jamesqf - you're not buy things by weight, but by mass ;)

@Dominik Szymański: No, I am buying by weight.  For instance, if I get some produce and load it on a typical grocery scale, that scale is measuring weight, not mass.  It's just that on Earth, the two happen to be equivalent.

@xxavier: See any dictionary for the definition of pedantic :-)

@jamesqf they absolutely are not equivalent. Weight is about 10 times bigger than mass on Earth, it just happens that scales are giving you mass based on assumption of given gravity. Look up any physics grade school book. Gram (and pound if you wish so) are units of mass. Unit of weight is Newton.
Scale on different planet would be just plainly lying. 1 liter of water by definition contains (in normal conditions, that is temperature and pressure) 1kg of water. However if you put the same amount of water (1 liter) on the scale on mars it would show around 0.39kg and that would be a lie.

@DominikSzymański You can't compare mass with force, since they are different things.

Common usage aside, if you want your SI calculations to work then you have to use Newtons as a unit of force


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What are the FAA regs regarding landing a helicopter off airport?
I have been mulling over getting a Mosquito helicopter, the experimental version with the uprated gas or turbine engine and larger fuel capacity. It's unwise to try to fly any of the Mosquitos without at least a rotary wing solo ticket, and if you have that, you may as well get the better version so you can log the time, and take advantage of the greater range and power. This leads to an interesting question:
What are the regulations regarding where a helicopter can be landed? Obviously, it can be landed in a lot of places, but what does officialdom have to say about where a helicopter can and cannot be landed?

Interesting question indeed. It could be extended to non-commercial aircraft as well, is it possible to land some kind of aircraft in your property, for those who have a big property.

oops, didn't see that one.

I am inclined to leave this open as not a duplicate because: A this question also appears to incorporate the ultralight question, and B the related question does not ask about FAA regulations as specifically, nor are any pertinent FAA regulations cited in the answers. In those answers, the only correct information offered is along the lines of "See state or local regulations".

There are no Federal Laws prohibiting landing rotor craft off field.
14 CFR 91.119

(1) A helicopter may be operated at less than the minimums prescribed in paragraph (b) or (c) of this section, provided each person operating the helicopter complies with any routes or altitudes specifically prescribed for helicopters by the FAA;

Part B and C, for reference

Except when necessary for takeoff or landing, no person may operate an aircraft below the following altitudes:
(b) Over congested areas. Over any congested area of a city, town, or settlement, or over any open air assembly of persons, an altitude of 1,000 feet above the highest obstacle within a horizontal radius of 2,000 feet of the aircraft.
(c) Over other than congested areas. An altitude of 500 feet above the surface, except over open water or sparsely populated areas. In those cases, the aircraft may not be operated closer than 500 feet to any person, vessel, vehicle, or structure.

However, state laws, local laws, zoning laws, etc. change from location to location. You need to make sure you plan where you are landing beforehand and check these laws.
For example, in the State of New Jersey, helicopters can only be landed at the following locations:

Permanent public use heliport or vertiport
Permanent special use or restricted use helistop or vertiport
Temporary helistop or vertiport

(Reference)


text:
If airliners and commercial planes have turbulence, can helicopters have turbulence?
I was watching a video of an V 22 (Osprey) taking off and it had that white cloudy stuff that looked like turbulence around the propellers. Can helicopters have turbulence too? I know an osprey is not a full helicopter nor airplane but it has some features of a helicopter.

I think you are asking about contrails or wingtip vortices, not turbulence. Turbulence is chaotic changes in air pressure which cause bumpiness when flying.

"white cloudy stuff that looked like turbulence". You can't see turbulence!

That 'cloudy stuff' that you say is not 'turbulence', but condensed water vapor, the condensation being due to the sudden cooling caused by the adiabatic expansion of the air due to the aerodynamic disturbance caused by a passing blade (or wing).
That condensation can also be observed as wakes in landing aircraft, and in stationary revving props. It appears when the sudden expansion takes place in humidity conditions close to saturation.

But yes, helicopters can experience turbulence just the same as an airplane. ;)

It's just as possible to get turb on a helicopter as on a fixed-wing aircraft: Turbulence is primarily caused by the characteristics of the air you're flying through, not the machine you're in.
That said, the rotors on a helicopter are reasonably good at damping turbulence (like the wings on an A380). As the V22 has smaller rotors it may be a little more bumpy than other - traditional - helicopters.

The super simple answer is that the helicopter moves through the air and with it.  It the air moves, then the helicopter will too. As to the condensation, this video shows it nicely.


text:
Statute miles to runway visual range
Is there a formula to converting statute miles to runway visual range? If so, what would it be? I'm seeing how 1 mile is equal to 5000 feet, but I am so confused because 1/2 of a mile is equal to 2,400 feet. I would think 1/2 would be 2500 feet. If it was half of a mile, you would take 5000 and divide that by .5 or 1/2 to get 2500. (I'm not a pilot yet)

1 statute mile is 5,280 feet, half of that is 2640. 1 nautical mile is 6076 feet, so I'm not sure where you are getting that half a mile is 2400 feet? RVR is usually already expressed in feet (or meters for our metric friends), I've never seen it shown in statute miles, mostly because RVR's over 6000' are not measured.

@RonBeyer I'm seeing it online if I look up "SM co converted to RVR" but thanks, I understand now.

Memorization.  This is the FAA definition:
Vis / RVR
1 SM / 5000
3/4 SM / 4000
1/2 SM / 2400
1/4 SM / 1600

This and a few more are spelled out in the Airman's Information Manual, TBL 5−4−1, RVR Value Conversions


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"Fundamental of Aerodynamics" 5th editon, example about oblique shock. Is the author's solution wrong?
I am reading "Fundamentals of Aerodynamics" 5th edition, J.D.Anderson. If you have the book, go to chapter 9: Oblique shock and expansion waves. Please look at example 9.2, page 618:
Consider a supersonic flow with M = 2, p = 1 atm, and T = 288 K. This flow is
deflected at a compression corner through 20◦. Calculate M, p, T, $p_0$, and $T_0$ behind the resulting oblique shock wave.(I only have problem with $p_0$).
Before see the solution, try to solve this example by yourself, you might need to see the appendices, here it is: https://drive.google.com/file/d/1MX6MASeP8pU-u_4MDjInfKIzPJfUfMaB/view?usp=sharing
And this is the solution from author:

The interesting thing here is: I was reading the note carefully, I still see something wrong here:

And this is my solution for $p_0,2$:
$p_2$ = 2.82 atm
From $M_2$ = 1.21 $\Rightarrow$ $p_0,2$/$p_2$ = 2.489 (isentropic flow, appendix A)
So $p_{0,2}$ = 2.82 x 2.489 = 7.019  atm
You may think the result is the same as the author's solution, but this is just coincident, look at example 9.5, page 620:
Consider a Mach 3 flow. It is desired to slow this flow to a subsonic speed. Consider two separate ways of achieving this: (1) the Mach 3 flow is slowed by passing directly through a normal shock wave; (2) the Mach 3 flow first passes through an oblique shock with a 40◦ wave angle, and then subsequently through a normal shock. These two cases are sketched in Figure 9.14. Calculate the ratio of the final total pressure values for the two cases, that is, the total pressure behind the normal shock for case 2 divided by the total pressure behind the normal shock for case 1. Comment on the significance of the result.

Again, try to solve this example by your own. Author's solution:

My solution:
Case 1: same as author's solution
Case 2:
$M_{n,1}$ = 1.93 $\Rightarrow$ $p_2$/$p_1$ = 4.224 (1)
$M_1 = 3$ $\Rightarrow$ $p_{0,1}$/$p_1$ = 36.73  (2)
(1) and (2) $\Rightarrow$ $p_{0,1}$/$p_2$ = 36.73/4.224 (3)
$M_2 = 1.9$ $\Rightarrow$ $p_{0,2}$/$p_2$ = 6.701 (4)
(3) and (4) $\Rightarrow$ $p_{0,2}$/$p_{0,1}$ = 6.701x4.224/36.73 = 0.771 (different from author's solution $p_{0,2}$/$p_{0,1}$ = 0.7535
$p_{0,3}$/$p_{0,2}$ is same as author's solution

In both examples, you can see what is going on. Am I wrong or Anderson wrong ??
You might think: "Oh, Anderson said that changes across an oblique shock wave are
governed only by the component of velocity normal to the wave." But this holds only for p, T, $\rho$, $M_{n,1}$, $M_{n,2}$. For $p_{0,1}$ and $p_{0,2}$, we must consider the tangential components of velocities.

Great question, we definitely have a few guys around here who can help out!

I’d suggest contacting Dr Anderson directly with that question as he could probably address it best.

Continue reading, I have found the answer. Look at problem 9.3, page 663:

And here is the author's solution:

Great that you found the answer, and thanks for coming back and sharing the knowledge


text:
Circular arc airfoil wing: why does the wing nose down twist due to camber?
I am reading this paper:
"Computational Investigation of Microscale Shrouded Rotor
Aerodynamics in Hover" Vinod K. Lakshminarayan, James D. Baeder ; JOURNAL OF THE AMERICAN HELICOPTER SOCIETY56, 042002 (2011).
In part "Rotor and Shroud Configuration" (page 2), the paper said:

" The blades used have a 2:1 taper starting at the 60% span location with the leading edge remaining straight and the trailing edge tapering toward the leading edge; this results in about 11 degrees nose down twist in this region due to the camber. The radius and the chord of the blade are 121 mm and 25 mm, respectively, making the aspect ratio equal to 4.84. A circular arc airfoil with 10% camber, 2% thickness, and leading edge sharpened from the 8% chord location is used."

I thought taper has nothing to do with angle of attack of airfoil section and the chord lines of every section lie in the same plane. How could the wing nose down due to camber?


text:
How to scale the data of a wing aspect ratio to correspond to the case of another aspect ratio?
I am reading "Fundamental of Aerodynamics " by J.D.Anderson Fifth edition. If you got the book, go to chapter 5, page 440 and 441. If you don't have the book then here you go: https://avionicsengineering.files.wordpress.com/2016/11/john-d-anderson-jr-fundamentals-of-aerodynamics.pdf  (page 465 and 466 of the pdf file).
I don't know how to insert the respective values of $C_{D,2}$ and $AR_2$ from figure 5.21 into Equation (5.66). How can we obtain the graph in figure 5.22 ? I am stuck at this point, can you help me out ?

Very good book!

For reference, here is figure 5.21, which is originally from NACA report 116 (1921):

Each curve represents a wing with an aspect ratio as labeled. In modern terms, $C_a=100C_\mathrm{L}$ and $C_w=100C_\mathrm{D}$, where $C_\mathrm{L}$ is lift coefficient and $C_\mathrm{D}$ is drag coefficient.
Anderson derives eq. (5.66), which scales the drag coefficient of a given wing (subscript 2) to that of a wing with aspect ratio of 5 (subscript 1):
\begin{equation}
C_{D,1}=C_{D,2}+\frac{C_{L}^2}{\pi e}\Bigg(\frac{1}{5}-\frac{1}{\mathrm{AR}_2}\Bigg)
\end{equation}
The original NACA report states that this formula has "been found to hold for distributions of lift [that] do not deviate too much from elliptical ones, although strictly speaking [it applies] only to the latter." I'm not sure why Anderson doesn't include this important point, but the implication is that we can safely take $e=1$.
Now proceed as follows: Select a curve from figure 5.21 and plug the corresponding aspect ratio into eq. (5.66). Then pick any $(C_w,C_a)$ pair off the curve and plug those into the equation. Plot the resultant $C_{w,1}$ on a figure against the same $C_{a}$. Continue with as many coordinate pairs as you want from the same curve; plot them all with the same symbol. Then select another curve and repeat but using a different symbol. Include a legend. The result should look something like figure 5.22:


text:
Why does the rotor thrust decrease at smaller tip clearance for shrouded rotor?
Consider this system, rotor with shroud:

From this research, we see that: at the same rpm, the rotor thrust decreases,the shroud thrust increases as tip clearance decreases. But why does the rotor thrust decrease at smaller tip clearance? You can see author said at page 11:

"Note that the increased inflow in the plane of the rotor at smaller tip clearances causes a small reduction in the effective angle of attack seen by the rotor blade, which results in a small decrease in the rotor thrust."

I understand that the inflow increases that will causes a small reduction in the effective angle of attack seen by the rotor blade. But final question: why do we have increased inflow at smaller tip clearance?

"final question: why do we have increased inflow": When I read the excerpt and see your diagram, I understand that if you have a longer blade (resulting from a reduced clearance with constant shroud diameter), the rotor disk diameter is increased, hence the air flow mass (assuming the angular speed is kept constant). Am I wrong in my reading?

@mins actually, the blade diameter is kept consant  and the shroud diameter changes (the author used CFD and easily changed shroud diameter )

This is the reason why a jet engine needs a shroud around its core: to reduce blade tip loss. When you decrease the wingtip clearance, you are curbing the wingtip vortices or the wingtip pressure bleed, and in doing so you are making each blade more and more effective at approximating its ideal i.e. infinite span high-low pressure spatial distribution under and above it, and this more ideal pressure difference accelerates the airflow better, which results in more inflow, which reduces the angle of attack of each blade a little, but increases the overall thrust produced. Think of the shroud as a big end plate for each and every blade. Think of this setup as the fan in a high bypass ratio turbofan: less fan tip clearance = more thrust.
EDIT: I'm not sure the author understood it as I do in his paper. He never clearly explained it in his writing. Plus, just why is there so much lift distributed at the wingtip? I've never seen anyone design a propeller with so much bending moment. He is making a propeller much heavier than it really is because the propeller could be less rigid and lighter and still providing the same thrust. If you really want an efficient prop, put the majority of lift near the midsection, and let the lift taper when approaching the tip.
EDIT: See how shroud works on a bona fide jet engine compressor:

Could you please explain these two points in your answer? 1) "which reduces the angle of attack of each blade a little"  2) "If you really want an efficient prop, put the majority of lift near the midsection, and let the lift taper when approaching the tip."

@Hans 1) the angle of attack of each blade is determined by the linear velocity of the blade and incoming speed of air when the latter increases the angle of attack decreases 2) Google the lifting line theory.

If I understand you correctly, you are saying that in order to get the same amount of inflow at higher incoming air speed the blade is designed to have its angle of attack reduced, rather than that for a fixed design, when the air speed increases the blade angle of attack adapts and increases. Is it right?

@Hans no, no blade is designed for any fixed AoA. Blades are rotating wings, and wings naturally tolerate a certain range of AoA before separation and stalling. How can the AoA not change when inflow increases if RPM is held constant? Unless the air is compressed, and that is unlikely, as air is decompressed a bit before the prop.

You are saying the following: The inflow increases the downward air speed through the blade while the rotor speed is held constant, the net direction of the air relative to the blade is turned more downwards compared to that at lower downward airflow speed, therefore the angle of attack of the blade relative to the net air flow direction decreases. Is it right?


text:
flow rate through a propeller disc
Let's take the case of a stock Cessna 150, chained to the ground, standard day conditions. The engine is running at full power (say 2500RPM) and the propeller is absorbing its full output. Is there a formula that allows the calculation of the volumetric rate of flow of air through the propeller disc for this static test case?

This answer might be relevant

thanks @pondlife, that is most excellent!

I am also surprised that you replied before our friend Peter Kaempf!

Chained to the ground, I didn't know that a C150 prop could absorb full output of the power plant.

aviation humor detected!


text:
How much horsepower is required to run the supercharger in an R-2800 engine?
How much horsepower was required to run the supercharger impeller in the R-2800 engine at cruise conditions, as used in the DC-6?

There may be a way to roughly calculate it based on fuel consumption and/or engine rpm.  Interesting that superchargers increased Horsepower output of an engine, but at the cost of higher fuel use.  The higher efficiencies (higher compression ratios) may have made it close to a "free lunch".

This is not an answer, just a comment. The answer can be reasonably determined for 65% power (assuming 100% of the sea-level rated power is available from boost at altitude) to be about 450 hp, with equivalent power also being provided to the shaft by the compounding turbine. Compounding increased engine thermal efficiency to about 36 pct; normal would be ~28-30 pct.  The assumed full rated hp at sea level is 2800 hp. The fuel savings from compounding is about 25 pct.

@nielsnielsen Are we good here?  Wow - talk about good fortune finding the source by Graham White noted farther below.  Maybe my previous impression was not a DC-6, but a DC-7, of which more sense can be made. I made an appropriate edit to the originally provided answer.  Evidently, although somewhat off base, the answer was apparently informative.   Best of regards.  -TP

@ThomasPerry, all good, thanks for your input, I appreciate it, answer accepted!- NN

For compounding see below.  The R-2800 was not compounded.  From information available in print, the power absorbed by the supercharger turbine was about 400 horsepower.  This is comparable to the power required by the compounding supercharger.  A comment below gives the source in print for the noted supercharger power requirement of 400 hp.  This was a very interesting question!

The caption explains these diagrams.  On the right, (b) shows the equivalent work produced by the supercharger and compounding turbine. The excess hatched area is equivalent to power produced at the propeller.  Since compounding will realize equivalent power at the propeller with 25 pct less fuel, the portion of the power at the propeller produced by the compounding is equivalent to the total shaft power from the turbine less that portion of the shaft power consumed by the compressor.  If we make the rough assumption that the area of the turbine curve (hatched area in b) is about twice that of the compressor curve, then the power consumed by the compressor would be equivalent to the half the power produced by the turbine.  The other half of the power from the turbine is given to the propeller.  In other words, if the engine is operated at 65 pct power, approximately one-fourth of that power is produced at the propeller by the turbine, and equivalent power is consumed by the compressor in boosting the engine.
If we do the math we will find the turbine produces about 910 hp, of which half is given to the propeller and half is given to the compressor.  These do not seem to be unreasonable approximations for the engine operating at 65 pct of its rated 2800 horsepower.
Source of the illustration -
Dommasch, et. al, Airplane Aerodynamics, 3rd ed. 1961, or 4th ed. 1967,  Pitman Publishing, New York.

But the R-2800 was not compounded, It had a single-stage, 2-speed supercharger driven off the crankshaft. It is the power consumed by that supercharger that I am interested in. -NN

Sorry, Niels.  If the answer is wrong, I'll take it down.  I do remember asking about that first time I saw a DC-6, what kind of engines were on that plane, and was told they were compound.  Ummm... yeh. I know; 68 yrs ago, now... Otherwise, I do not know the power absorbed by the direct-drive compressor.

OK - I found your answer! Up to 400 hp is necessary to drive the supercharger in order to produce the necessary mass air flow and manifold pressure. source: Graham White, 2001. R-2800: Pratt and Whitney's Dependable Masterpiece. Society of Automotive Engineers, Warrendale, PA. p. 89.

Hum BABY! 400HP to spin that puppy! Time for my next question! -NN

As compounding is not relevant to this question, I think it should be removed from this answer.

@Jpe61 Removal of information regarding compounding had been considered, but at that time there were already votes thereby indicating that some people found the presented information was informative.  Consequently, the decision was made to keep the total aspect of the answer when on-target and relevant information was discovered & presented.  Further, the apparent blower power requirement is about the same whether compounded or by direct drive.  For direct drive, this is about 25% of the rated power setting. For the R-2800 at cruise, 2100hp x 0.65 x .25 = 390hp. This is, however, unverified.

Ok, I see. However compound charger and simple crank driven supercharger are two very different setups, even though the diagrams look alike.

@Jpe61 Look only at diagram (b). The hatched area is power augmentation from the power-recovery turbine in the exhaust flow.  Without power recovery, power required by the supercharger is area A',B',C',D'. With power recovery, power required by the supercharger is a "free lunch," as R. DiGiovanni would say, with the surplus being supplied directly to the engine [1]. The power-recovery turbine is connected directly to the crankshaft & provides direct power to the supercharger as well as the propeller. [1]: https://babel.hathitrust.org/cgi/pt?id=mdp.39015002085887&seq=274

Yes. And if the FI system is a crank driven one, it will take just as much power to run with or without compounding. They are separate systems per se.


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What's up with the long exhaust stack I saw on a Cessna Caravan?
Our local airport is regularly visited by Cessna caravans flying cargo. This week I saw one parked outside the FBO building that had its exhaust stack bent down to lead under the cowling, where it was connected to a duct suspended from the underside of the fuselage. The duct was perhaps a foot in diameter and ran all the way back to where the tailcone began. Every other Caravan I have ever seen has the exhaust stack exiting the cowling on one side and then terminating. What's up with the ducted exhaust?

Caravans or Grand Caravans? Is this particular plane a turboprop, I would imagine so judging by the size of the exhaust?

https://www.thedrive.com/the-war-zone/14088/this-unique-epa-plane-is-monitoring-toxic-waste-dumps-around-houston   Was it this plane?

That's an answer.  It's to clear the exhaust of atmospheric sampling intakes.

@md88fan yes yes that's it!!!

Holy cow! That’s a lucky catch. Glad I could help.

Somebody write an answer.  @md88fan go for it.

The aircraft you saw, "ASPECT," is a contractor operated Cessna Grand Caravan used by the EPA to monitor sites of active or possible CBRN disasters.

The reason for the long exhaust stack is simply to keep exhaust gasses from interfering with the sensors on board. As I write this, the aircraft is flying over New Orleans, likely assessing damage from Hurricane Ida:

Source: flightaware.com; Sep 5th, 2021
The mission is similar, though not identical, to the Bell 412s operated by the Dept of State, equipped with sensors to measure radiation- baselines are established for cities and these helicopters can be deployed in the unlikely scenario that radiological material is stolen or a nuclear/radiological weapon is suspected to be nearby. Hope this helps!
Update: per StephenS, this is also common on aerial survey aircraft, but since it has been identified as the ASPECT aircraft, the above remains true.

Yes it does! this was great.

This is also common on aerial survey planes to keep hot, dirty exhaust away from the expensive cameras.